Logistic growth and phase lines, question 10

Topic: Logistic growth and phase lines
Difficulty: Capstone
Suggested time: 12 min
Marks: 6
Theory: UBC Math 100 §13.1Key results

A protein switches on its own production. Its concentration $c(t) \ge 0$ satisfies the equation below, where the first term is production and $-c$ is decay.

\frac{dc}{dt} = \frac{10c^{2}}{9 + c^{2}} - c

(a) Find every steady state. (b) Draw the state-space diagram (phase line) for c ≥ 0 and classify each steady state as stable or unstable. (c) Describe what happens in the long run if c(0) = 0.5, and if c(0) = 3.

Your answer

Theory UBC Math 100 §13.1Key results

Worked solution

Strategy

Combine the right side into one fraction and factor it. Its zeros are the steady states, and its sign between them gives the arrows on the state-space diagram, which settle both stability and the long run from each start.

Working

$\frac{dc}{dt} = \frac{10c^{2} - 9c - c^{3}}{9 + c^{2}} = \frac{-c\,(c - 1)(c - 9)}{9 + c^{2}}$
One factored fraction shows every zero and every sign change.
$\frac{dc}{dt} = 0 \iff c = 0,\ 1 \text{ or } 9$
The denominator 9 + c² is never zero, so these are all the steady states.
$c = 0.5:\ \frac{dc}{dt} < 0 \qquad c = 3:\ \frac{dc}{dt} = \frac{36}{18} = 2 > 0 \qquad c = 10:\ \frac{dc}{dt} < 0$
The sign can change only at a steady state, so one test value per interval fixes it.
$\underset{\text{stable}}{0}\ \longleftarrow\ \underset{\text{unstable}}{1}\ \longrightarrow\ \underset{\text{stable}}{9}\ \longleftarrow$
The arrows point at 0 from above and at 9 from both sides, and away from 1 on both sides.
$c(0) = 0.5:\ c(t) \to 0, \qquad c(0) = 3:\ c(t) \to 9$
Below the threshold 1 the protein dies away; above it, production wins and the switch turns on.

Answer

\text{Steady states } 0 \text{ (stable)},\quad 1 \text{ (unstable)},\quad 9 \text{ (stable)};\quad c(0) = 0.5:\ c \to 0,\quad c(0) = 3:\ c \to 9

What earns the marks

Sets the right-hand side to zero without dividing by c, and finds c = 0 together with the roots of c² − 10c + 9 = 0: c = 1 and c = 9.
Writes the right side as −c(c − 1)(c − 9)/(9 + c²), or tests one value in each interval, and finds dc/dt < 0 on (0, 1), dc/dt > 0 on (1, 9) and dc/dt < 0 for c > 9.
Draws arrows on the c-axis from those signs, and classifies c = 0 and c = 9 as stable and c = 1 as unstable.
Concludes that from c(0) = 0.5 the concentration decreases towards 0, and from c(0) = 3 it increases towards 9, neither crossing a steady state.

Check

Production 10c²/(9 + c²) never reaches 10, so decay c wins for every c ≥ 10, and at c = 0.5 production is only about 0.27, below the decay 0.5.

Where this goes wrong

Dividing 10c²/(9 + c²) = c by c and losing the steady state c = 0, or assuming every solution ends at the largest steady state. The unstable steady state c = 1 is a threshold: below it the protein dies away, above it the concentration rises to 9.

Leans on

Signs of a factored function, from Curve sketching

Try a similar question

The differential equation below has steady states

y = 1

and

y = 4

. Which statement describes their stability?

\frac{dy}{dt} = (y - 1)(y - 4)

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Logistic growth and phase lines practice question 10